Elements of Classical Field Theory

C6, HT 2017

Uli Haischa

aRudolf Peierls Centre for Theoretical PhysicsUniversity of Oxford

OX1 3PN Oxford, United Kingdom

Please send corrections to u.haisch1@physics.ox.ac.uk.

The parts of the script that are colored blue are not covered in the lecture. They will howeverbe discussed (at least in parts) in the exercises.

1 Classical Field Theory

In this part of the lecture we will discuss various aspects of classical fields. We will cover onlythe bare minimum ground necessary before turning to the quantum theory, and will returnto classical field theory at several later stages in the course when we need to introduce newconcepts or ideas.

1.1 Lorentz Group

The Lorentz group L is of fundamental importance for the construction of relativistic fieldtheories, since L is associated to the symmetry of 4-dimensional space-time.

Using the Minkowski metric η = diag (1,−1,−1,−1), the Lorentz group L consists of thereal 4× 4 matrices Λ that satisfy

η = ΛT ηΛ , (1.1)

which written in components (µ, ν, ρ, σ = 0, 1, 2, 3) takes the form

ηµν = ησρΛµρΛν

σ . (1.2)

It is readily seen that the transformations

xµ → Λµνx

ν , (1.3)

with Λ satisfying (1.1) leave the distance ds2 invariant. Setting for simplicity the speed oflight c to 1, one has

ds2 = ηµνdxµdxν → ηµνΛµ

ρdxρΛν

σdxσ = ηρσdx

ρdxσ = ds2 . (1.4)

1

4.1. SYMMETRIES 43

det(!) !00 name contains given by

+1 ! 1 L!+ 14 L!

+

+1 " #1 L"+ PT PTL!

+

#1 ! 1 L!# P PL!

+

#1 " #1 L"# T TL!

+

Table 4.1: The four disconnected components of the Lorentz group. The union L+ = L!+ $ L"

+ is also called theproper Lorentz group andL! = L!

+$L!# is called the orthochronosLorentz group (as it consists of transformations

preserving the direction of time). L!+ is called the proper orthochronos Lorentz group.

O satisfying OT O = 1 and det(O) = 1. Writing O = 1 + iT with (purely imaginary) generators T , the relationOT O = 1 implies T = T † and, hence, that the Lie-algebra of SO(3) consists of 3 % 3 anti-symmetric matrices(multiplied by i). A basis for this Lie algebra is provided by the three matrices Ti defined by

(Ti)jk = #i!ijk , (4.16)

which satisfy the commutation relations[Ti, Tj] = i!ijkTk . (4.17)

These are the same commutation relations as in Eq. (4.13) and, hence, the Ti form a three-dimensional (irreducible)representation of (the Lie algebra of) SU(2). This representation must fit into the above classification of SU(2)representations by an integer or half-integer number j and, simply on dimensional grounds, it has to be identifiedwith the j = 1 representation.

4.1.4 The Lorentz groupThe Lorentz group is of fundamental importance for the construction of field theories. It is the symmetry associatedto four-dimensional Lorentz space-time and should be respected by field theories formulated in Lorentz space-time.Let us begin by formally defining the Lorentz group. With the Lorentz metric " = diag(1, #1, #1, #1) the Lorentzgroup L consists of real 4 % 4 matrices ! satisfying

!T "! = " . (4.18)

Special Lorentz transformations are the identity 14, parityP = diag(1, #1, #1, #1), time inversionT = diag(#1, 1, 1, 1)and the product PT = #14. We note that the four matrices {14, P, T, PT } form a finite sub-group of the Lorentzgroup. By taking the determinant of the defining relation (4.18) we immediately learn that det(!) = ±1 for allLorentz transformations. Further, if we write out Eq. (4.18) with indices

"µ!!µ"!

!# = ""# (4.19)

and focus on the component # = $ = 0 we conclude that (!00)

2 = 1 +!

i(!i0)

2 ! 1, so either !00 ! 1

or !00 " #1. This sign choice for !0

0 combined with the choice for det(!) leads to four classes of Lorentztransformations which are summarised in Table 4.1. Also note that the Lorentz group contains three-dimensionalrotations since matrices of the form

! =

"1 00 O

#(4.20)

satisfy the relation (4.18) and are hence special Lorentz transformations as long as O satisfies OT O = 13.To find the Lie algebra of the Lorentz group we write! = 14+iT + . . . with purely imaginary 4%4 generators

T . The defining relation (4.18) then implies for the generators that T = #"T T ", so T must be anti-symmetricin the space-space components and symmetric in the space-time components. The space of such matrices is six-dimensional and spanned by

Ji =

"0 00 Ti

#, K1 =

$%%&

0 i 0 0i 0 0 00 0 0 00 0 0 0

'(() , K2 =

$%%&

0 0 i 00 0 0 0i 0 0 00 0 0 0

'(() , K3 =

$%%&

0 0 0 i0 0 0 00 0 0 0i 0 0 0

'(() , (4.21)

Figure 1.1: The four disconnected components of the Lorentz group.

The Lorentz transformations (LTs) (1.3) are therefore consistent with the postulate of specialrelativity that tells us that the speed of light is the same in all inertial frames.

Let us examine the consequences of (1.1). First, we take its determinant

det (η) = det(ΛT)

det (η) det (Λ) , (1.5)

from which we deduce thatdet (Λ) = ±1 . (1.6)

The case of det (Λ) = +1 (−1) corresponds to proper (improper) LTs and the associatedsubgroup is L+ (L−). This implies that parity or space-inversion P = diag (1,−1,−1,−1) aswell as time-reversal T = diag (−1, 1, 1, 1) are improper LTs and as such part of L−. Second,we look at the component η00, in which case one finds from (1.2) the relation

1 = ησρΛ0ρΛ0

σ =(Λ0

0

)2 −∑

i=1,2,3

(Λi

0

)2. (1.7)

It follows that|Λ0

0| ≥ 1 . (1.8)

When Λ00 ≥ 1 the LT is said to be orthochronous and part of L↑, while Λ0

0 ≤ −1 gives anon-orthochronous LT which belongs to L↓. In consequence, the Lorentz group consists outof four classes of LTs as illustrated in Figure 1.1

Let us have a look at some simple example of LTs. A rotation by the angle θ about thez-axis and a boost by v < 1 along the x-axis

Λµν =

1 0 0 0

0 cos θ − sin θ 0

0 sin θ cos θ 0

0 0 0 1

, Λµ

ν =

γ −γv 0 0

−γv γ 0 0

0 0 1 0

0 0 0 1

, (1.9)

with γ = (1− v2)−1/2 are part of the Lorentz group. In fact, any 3-dimensional rotation

Λµν =

(1 0

0 O

), (1.10)

2

44 CHAPTER 4. CLASSICAL FIELD THEORY

(j+, j!) dimension name symbol(0, 0) 1 scalar !(1/2, 0) 2 left-handed Weyl spinor "L

(0, 1/2) 2 right-handedWeyl spinor "R

(1/2, 0) ! (0, 1/2) 4 Dirac spinor #(1/2, 1/2) 4 vector Aµ

Table 4.2: Low-dimensional representations of the Lorentz group.

where Ti are the generators (4.16) of the rotation group. Given the embedding (4.20) of the rotation group intothe Lorentz group the appearance of the Ti should not come as a surprise. It is straightforward to work out thecommutation relations

[Ji, Jj ] = i$ijkJk , [Ki, Kj ] = "i$ijkJk , [Ji, Kj] = i$ijkKk . (4.22)

The above matrices can also be written in a four-dimensional covariant form by introducing six 4#4matrices %µ! ,labelled by two anti-symmetric four-indices and defined by

(%µ!)"# = i(&"

µ&!# " &µ#&"!) . (4.23)

By explicit computation one finds that Ji = 12$ijk%jk andKi = %0i. Introducing six independent parameters $µ! ,

labelled by an anti-symmetric pair of indices, a Lorentz transformation close to the identity can be written as

!"# $ '"

# " i

2$µ!(%µ!)"

# = '"# + $"

#; . (4.24)

The commutation relations (4.22) for the Lorentz group are very close to the ones for SU(2) in Eq. (4.13). Thisanalogy can be made even more explicit by introducing a new basis of generators

J±i =

1

2(Ji ± iKi) . (4.25)

In terms of these generators, the algebra (4.22) takes the form

[J±i , J±

j ] = i$ijkJ±k , [J+

i , J!j ] = 0 , (4.26)

that is, precisely the form of two copies (a direct sum) of two SU(2) Lie-algebras. Irreducible representations of theLorentz group can therefore be labelled by a pair (j+, j!) of two spins and the dimension of these representationsis (2j++1)(2j!+1). A list of a few low-dimensional Lorentz-group representations is provided in Table 4.2. Fieldtheories in Minkowski space usually require Lorentz invariance and, hence, the Lorentz group is of fundamentalimportance for such theories. Since it is related to the symmetries of space-time it is often also referred as externalsymmetry of the theory. The classification of Lorentz group representations in Table 4.2 provides us with objectswhich transform in a definite way under Lorentz transformations and, hence, are the main building blocks ofsuch field theories. In these lectures, we will not consider spinors in any more detail but focus on scalar fields!, transforming as singlets, ! % ! under the Lorentz group, and vector fields Aµ, transforming as vectors,Aµ % !µ

!A! .

4.2 General classical field theory4.2.1 Lagrangians and Hamiltonians in classical field theoryIn this subsection, we develop the general Lagrangian and Hamiltonian formalism for classical field theories.This formalism is in many ways analogous to the Lagrangian and Hamiltonian formulation of classical mechan-ics. In classical mechanics the main objects are the generalised coordinates qi = qi(t) which depend on timeonly. Here, we will instead be dealing with fields, that is functions of four-dimensional coordinates x = (xµ)on Minkowski space. Lorentz indices µ, (, · · · = 0, 1, 2, 3 are lowered and raised with the Minkowski metric(&µ!) = diag(1, "1, "1, "1) and its inverse &µ! . For now we will work with a generic set of fields !a = !a(x)before discussing scalar and vector fields in more detailed in subsequent sections. Recall that the Lagrangian in

Figure 1.2: Low-dimensional representations of the Lorentz group.

leaves ds2 invariant, since one has OTO = OOT = 13 by definition, if O is an element of the3-dimensional rotation group SO(3).1

Any LT can be decomposed as the product of a rotation, a boost, space-inversion P ,and time-reversal T . Let us concentrate on the continuous transformations. Since thereare three rotations and three boosts, one for each space direction, the continuous LTs aredescribed in terms of six parameters. To find the corresponding six generators, i.e., a basis oftransformation matrices that describes infinitesimal rotations and boosts, we write

Λ = 14 + iT , (1.11)

where T are purely imaginary 4 × 4 matrices. Inserting this linearized LT into the definingrelation (1.1), implies

T = −ηT T η . (1.12)

This tells us that the generators T must be anti-symmetric in the space-space components, butsymmetric in the space-time components. The space of such matrices has indeed dimensionsix and is spanned by the set (i = 1, 2, 3)

Ji =

(0 0

0 Ti

),

K1 =

0 i 0 0

i 0 0 0

0 0 0 0

0 0 0 0

, K2 =

0 0 i 0

0 0 0 0

i 0 0 0

0 0 0 0

, K3 =

0 0 0 i

0 0 0 0

0 0 0 0

i 0 0 0

.

(1.13)

Here (Ti)jk = −iεijk with εijk the fully anti-symmetric Levi-Civita tensor (ε123 = +1) are thegenerators of SO(3). It follows that the matrices Ji (Ki) generate rotations (boosts).

It is a matter of simply algebra to work out the commutation relations of the generatorsJi and Ki of the Lorentz group. One obtains

[Ji, Jj] = iεijkJk , [Ki, Kj] = −iεijkJk , [Ji, Kj] = iεijkKk , (1.14)

1The group of 3-dimensional orthogonal matrices is denoted by O(3), while its subgroup of matrices withdeterminant +1 is called the special orthogonal group SO(3).

3

where [a, b] = ab − ba is the usual commutator. These relations look very similar to thecommutation relations of angular momentum or spin that you should know from quantummechanics (QM). To make the analogy between the Lorentz group and the Lie group SU(2)even more explicit, we introduce the following new basis of generators

J±i =1

2(Ji ± iKi) . (1.15)

In terms of these generators, the Lorentz Lie algebra (1.14) takes the form

[J±i , J±j ] = iεijkJ

±k , [J±i , J

∓j ] = 0 . (1.16)

This means that J+i and J−i independently obey the Lie algebra of SU(2). By analogy to the

spin quantum number, we can therefore introduce a pair (j+, j−) of two spins that characterizethe possible representations of the Lorentz group. The states within a representation arefurther distinguished by the eigenvalues of J+

3 and J−3 , which can take the values m+ =−j+,−j+ + 1, . . . , j+ − 1, j+ and m− = −j−,−j− + 1, . . . , j− − 1, j−. The dimension, i.e., thenumber of distinct states in a given representation (j+, j−) is hence (2j+ + 1) (2j− + 1).

A list of a few low-dimensional Lorentz-group representations is provided in Figure 1.2.Field theories in Minkowski space usually require Lorentz invariance and, hence, the Lorentzgroup is of fundamental importance for such theories. Since it is related to the symmetries ofspace-time it is often also referred as external symmetry of the theory. The classification ofLorentz group representations provides us with objects which transform in a definite way underLTs and thus these objects are the main building blocks of such field theories. In what followswe will not deal with the spinor fields χL,R and ψ that describe fermions (quarks and leptons).Our focus will be on scalar φ and vector Aµ fields, which have very simple transformationproperties under the Lorentz group:

φ→ φ , Aµ → ΛµνAν . (1.17)

1.2 Dynamics of Fields

A field is a quantity defined at every space-time point x = (t,x). While classical particlemechanics deals with a finite number of generalized coordinates qa(t), indexed by a label a, infield theory we are interested in the dynamics of fields

φa(t,x) , (1.18)

where both a and x are considered as labels. We are hence dealing with an infinite number ofdegrees of freedom (dofs), at least one for each point x in space. Notice that the concept ofposition has been relegated from a dynamical variable in particle mechanics to a mere labelin field theory.

Lagrangian and Action

The dynamics of the fields is governed by the Lagrangian. In all the systems we will studyin this course, the Lagrangian is a function of the fields φa and their derivatives ∂µφa,

2 and

2If there is no (or only little) room for confusion, we will often drop the arguments of functions and writeφa = φa(x) etc. to keep the notation short.

4

given by

L(t) =

∫d3xL(φa, ∂µφa) , (1.19)

where the official name for L is Lagrangian density. Like everybody else we will, however, sim-ply call it Lagrangian from now on. For any time interval t ∈ [t1, t2], the action correspondingto (1.19) reads

S =

∫ t2

t1

dt

∫d3xL =

∫d4xL . (1.20)

Recall that in classical mechanics L depends only on qa and qa, but not on the second timederivatives of the generalized coordinates. In field theory we similarly restrict to LagrangiansL depending on φa and φa. Furthermore, with an eye on Lorentz invariance, we will onlyconsider Lagrangians depending on ∇φa and not higher derivatives.

Notice that employing c = ~ = 1, i.e., working with natural units, the dimension ofthe action is [S] = 0. With (1.20) and [d4x] = −4, it follows that the Lagrangian mustnecessarily have [L] = 4. Other objects that we will use frequently to construct Lagrangiansare derivatives, masses, couplings, and most importantly fields. The dimensions of the formertwo objects are [∂µ] = 1 and [m] = 1, while the dimensions of the latter two quantities dependon the specific type of coupling and field one considers. We therefore postpone the discussionof the mass dimension of couplings and fields to the point when we meet the relevant buildingblocks.

Principle of Least Action

The dynamical behavior of fields can be determined by the principle of least action. Thisprinciple states that when a system evolves from one given configuration to another betweentimes t1 and t2 it does so along the “path” in configuration space for which the action is anextremum (usually a minimum) and hence satisfies δS = 0. This condition can be rewritten,using partial integration, as follows

δS =

∫d4x

{∂L∂φa

δφa +∂L

∂(∂µφa)δ(∂µφa)

}

=

∫d4x

{[∂L∂φa− ∂µ

(∂L

∂(∂µφa)

)]δφa + ∂µ

(∂L

∂(∂µφa)δφa

)}= 0 .

(1.21)

The last term is a total derivative and vanishes for any δφa that decays at spatial infinity andobeys δφa(t1,x) = δφa(t2,x) = 0. For all such paths, we obtain the Euler-Lagrange equationsof motion (EOMs) for the fields φa, namely

∂µ

(∂L

∂(∂µφa)

)− ∂L∂φa

= 0 . (1.22)

Hamiltonian Formalism

The link between the Lagrangian formalism and the quantum theory goes via the path integral.While this is a powerful formalism, we will for the time being use canonical quantization, since

5

it makes the transition to QM easier. For this we need the Hamiltonian formalism of fieldtheory. We start by defining the momentum density πa(x) conjugate to φa(x),

πa =∂L∂φa

. (1.23)

In terms of πa, φa, and L the Hamiltonian density is given by

H = πaφa − L , (1.24)

where, as in classical mechanics, we have eliminated φa in favor of πa everywhere in H. TheHamiltonian then simply takes the form

H =

∫d3xH . (1.25)

1.3 Noether’s Theorem

The role of symmetries in field theory is possibly even more important than in particle me-chanics. There are Lorentz symmetry, internal symmetries, gauge symmetries, supersymme-tries, etc. We start here by recasting Noether’s theorem in a field theoretic framework.

Currents and Charges

Noether’s theorem states that every continuous symmetry of the Lagrangian gives rise to aconserved current Jµ(x), so that the EOMs (1.22) imply

∂µJµ = 0 , (1.26)

or in components dJ0/dt+∇ ·J = 0. To every conserved current there exists also a conserved(global) charge Q, i.e., a physical quantity which stays the same value at all times, defined as

Q =

∫

R3

d3x J0 . (1.27)

The latter statement is readily shown by taking the time derivative of Q,

dQ

dt=

∫

R3

d3xdJ0

dt= −

∫

R3

d3x∇ · J , (1.28)

which is zero, if one assumes that J falls off sufficiently fast as |x| → ∞. Notice, however, thatthe existence of the conserved current J is much stronger than the existence of the (global)charge Q, because it implies that charge is in fact conserved locally. To see this, we define thecharge in a finite volume V by

QV =

∫

V

d3x J0 . (1.29)

6

Repeating the above analysis, we find

dQV

dt= −

∫

V

d3x∇ · J = −∮

S

dS · J , (1.30)

where S denotes the area bounding V , dS is a shorthand for n dS with n being the outwardpointing unit normal vector of the boundary S, and we have used Gauss’ theorem. In physicalterms the result means that any charge leaving V must be accounted for by a flow of thecurrent 3-vector J out of the volume. This kind of local conservation law of charge holds inany local field theory.

Proof of Theorem

In order to prove Noether’s theorem, we’ll consider infinitesimal transformations. This isalways possible in the case of a continuous symmetry. We say that δφa is a symmetry of thetheory, if the Lagrangian changes by a total derivative

δL(φa) = ∂µJ µ(φa) , (1.31)

for a set of functions J µ. We then consider the transformation of L under an arbitrary changeof field δφa. Glancing at (1.21) tells us that in this case

δL =

[∂L∂φa− ∂µ

(∂L

∂(∂µφa)

)]δφa + ∂µ

(∂L

∂(∂µφa)δφa

). (1.32)

When the EOMs are satisfied than the term in square bracket vanishes so that we are simplyleft with the total derivative term. For a symmetry transformation satisfying (1.31), therelation (1.32) hence takes the form

∂µJ µ = δL = ∂µ

(∂L

∂(∂µφa)δφa

), (1.33)

or simply ∂µJµ = 0 with

Jµ =∂L

∂(∂µφa)δφa − J µ , (1.34)

which completes the proof. Notice that if the Lagrangian is invariant under the infinitesimaltransformation δφa, i.e., δL = 0, then J µ = 0 and Jµ contains only the first term on theright-hand side of (1.34).

We stress that that our proof only goes through for continuous transformations for whichthere exists a choice of the transformation parameters resulting in a unit transformation, i.e.,no transformation. An example is a Lorentz boost with some velocity v, where for v = 0 thecoordinates x remain unchanged. There are examples of symmetry transformations where thisdoes not occur. E.g., a parity transformation P does not have this property, and Noether’stheorem is not applicable then.

7

Energy-Momentum Tensor

Recall that in classic particle mechanics, spatial translation invariance gives rise to the con-servation of momentum, while invariance under time translations is responsible for the con-servation of energy. What happens in classical field theory? To figure it out, let’s have a lookat infinitesimal translations

xν → xν − εν =⇒ φa(x)→ φa(x+ ε) = φa(x) + εν∂νφa(x) , (1.35)

where the sign in the field transformation is plus, instead of minus, because we are doing anactive, as opposed to passive, transformation. If the Lagrangian does not explicitly dependon x but only through φa(x) (which will always be the case in the Lagrangians discussed inthe course), the Lagrangian transforms under the infinitesimal translation as

L → L+ εν∂νL . (1.36)

Since the change in L is a total derivative, we can invoke Noether’s theorem which givesus four conserved currents T µν = (Jµ)ν one for each of the translations εν (ν = 0, 1, 2, 3).From (1.34) and (1.35) we readily read off the explicit expressions for T µν ,

T µν =∂L

∂(∂µφa)∂νφa − δµνL . (1.37)

This quantity is called the energy-momentum (or stress-energy) tensor. It has dimension[T µν ] = 4 and satisfies

∂µTµν = 0 . (1.38)

The four “conserved charges” are (µ = 0, 1, 2, 3)

P µ =

∫d3xT 0µ , (1.39)

Specifically, the “time component” of P µ is

P 0 =

∫d3xT 00 =

∫d3x

(πaφa − L

), (1.40)

which (looking at (1.24) and (1.25)) is nothing but the Hamiltonian H. We thus concludethat the charge P 0 is the total energy of the field configuration, and it is conserved. In fieldstheory, energy conservation is thus a pure consequence of time translation symmetry, like itwas in particle mechanics. Similarly, we can identify the charges P i (i = 1, 2, 3),

P i =

∫d3xT 0i = −

∫d3x πa∂iφa , (1.41)

as the momentum components of the field configuration in the three space directions, and theyare of course also conserved.

8

1.4 Scalar Field Theories

In the following we will apply the formalism described above to the simplest class of Lorentzinvariant field theories, involving only fields that transform trivially under the Lorentz group.We start by considering the case of a single real scalar field.

A Single Real Scalar Field

The Lagrangian density of our real scalar field φ is given by

L =1

2(∂µφ)2 − V (φ) , (1.42)

where the first term is referred to as kinetic energy (or kinetic term), while the second is knownas the scalar potential. Notice that φ corresponds to the (j+, j−) = (0, 0) representation of theLorentz group. The dimension of the real scalar field is obviously [φ] = 1.

The scalar potential can be written as

V =1

2m2φ2 +

∞∑

n=3

λnn!φn . (1.43)

Here m is the mass of φ (this will become clear later on) and the coefficients λn are calledcoupling constants. Dimensional analysis tells us that

[m] = 1 , [λn] = 4− n . (1.44)

The coupling terms in (1.43) fall into three different categories. First, dimension-threeterms with [λ3] = 1. For such terms, we can define a dimensionless parameter λ3/E, whereE has dimension of mass and represents the energy scale of the process of interest. Thismeans that ∆L3 = λ3 φ

3/(3!) is a small perturbation for high energies, i.e., E � λ3, but abig one at low energies, i.e., E � λ3. Such terms are called relevant, because they becomeand are most relevant at low energies which, after all, is where most of the physics that weexperience lies. In a relativistic quantum field theory (QFT), we have E > m, which meansthat we can always make this sort of perturbations small by taking λ3 � m. Second, termsof dimension four with [λ4] = 0. E.g., ∆L4 = λ4 φ

4/(4!). Such terms are small if λ4 � 1 andare called marginal. Third, operators with dimension of higher than four, having [λn] < 0. Inthis case the appropriate dimensionless parameters is (λnE

n−4) and terms ∆Ln = λn φn/(n!)

with n ≥ 5 are small (large) at low (high) energies. Such contributions are called irrelevant,since in daily life, meaning En−4 � λn, these terms do not matter.

As we will see later, it is typically impossible to avoid high-energy processes in QFT. Wehence might expect problems with irrelevant terms (or operators) that become important athigh energies. Indeed, these operators lead to non-renormalizable QFTs in which one cannotmake sense of the infinities at arbitrarily high energies. This does not mean that these theoriesare useless, it just means that they become incomplete at some energy scale and need to beembedded into an appropriate complete theory aka an ultraviolet (UV) completion. Let mealso add that the above naive assignment of relevant, marginal, and irrelevant operators is notalways carved in stone, since quantum corrections can sometimes change the character of anoperator.

9

Low-Energy Description

In typical applications of QFT only the relevant and marginal couplings are important. Thisis due to the fact that the irrelevant couplings become small at low energies, as we have seenabove. In practice this saves us, since instead of considering the infinite number of couplingterms in (1.43), only a handful are actually needed. E.g., in the case of the real scalar field φdescribed earlier, we only have to take into account two operators, namely ∆L3 = λ3 φ

3/(3!)and ∆L4 = λ4 φ

4/(4!), in the low-energy limit.Let us have a closer look at this issue. Suppose that at some day we discover the true

superduper theory aka the TOE that describes the world at very high energy scales. Whateverthis scale is, let’s call it Λ. Since it is an energy scale, we obviously have [Λ] = 1. What wewant to understand are the laws of physics at energy scales E that we can probe directlyin a laboratory, which given today’s standards, means E � Λ. Let us further suppose thatat energies of order E, the laws of physics are described by a real scalar field.3 This scalarfield will have some complicated coupling terms (1.43), where the precise form is dictated byall the stuff that is going on in the TOE. Can we get an idea about the interactions? Well,we can write our dimensionful coupling constants λn in terms of dimensionless couplings gn,multiplied by a suitable power of the relevant scale Λ,

λn =gn

Λn−4 . (1.45)

The exact values of the dimensionless couplings gn depend on the details of the TOE,4 so wehave to do some guesswork. Since the couplings gn are dimensionless, 1 looks like a pretty goodand somehow a natural guess. Since we are not completely sure, let’s say gn = O(1). Thismeans that in a laboratory with E � Λ the coupling terms ∆Ln = λnφ

n/(n!) of (1.43) will besuppressed by powers of (E/Λ)n−4 if n ≥ 5. Given the CERN Large Hadron Collider (LHC)energy of around 1 TeV, this is a suppression by many orders of magnitude. E.g., for Λ =MP = 1016 TeV corresponding to the Planck mass MP , one has E/Λ = 10−16. It is this simpleargument based on dimensional analysis that ensures that we need to focus only on the firstfew terms in the interactions, namely those that are relevant and marginal. It also means thatif we only have access to low-energy experiments, it is going to be very difficult to figure outthe precise nature of the TOE, because its effects are highly diluted except for the relevantand marginal interactions. Some people therefore call the superduper theory that everybodyis looking for, not TOE, but TOENAIL, which stands for “theory of everything not accessiblein laboratories”.

Klein-Gordon Equation

Applying the Euler-Lagrange equations to (1.42) leads to

∂µ∂µφ+∂V

∂φ= 0 . (1.46)

3Of course, we know that this assumption is plain wrong, since the standard model (SM) of particle physicsis a non-abelian gauge theory with chiral fermions, but the same argument applies in that case.

4If we would know the precise structure of the TOE we could, in fact, calculate the couplings gn.

10

The Laplacian in Minkowski space is sometimes denoted by � and ∂V/∂φ is commonly writtenas V ′. In this notation, the above equation reads �φ+ V ′ = 0.

Since for non-vanishing coefficients λn the solutions to (1.46) are difficult to find, let’ssimply ignore them for the time being and only consider the free case. The dynamics of sucha non-interacting massive real scalar field is encoded in the famous Klein-Gordon equation:

(� +m2

)φ = 0 . (1.47)

The corresponding Lagrangian is of course

L =1

2(∂µφ)2 − 1

2m2φ2 . (1.48)

In order to find the solutions to the classical Klein-Gordon equation (1.48), we only haveto Fourier transform the field φ,

φ(t,x) =

∫d3p

(2π)3eip·x φ(t,p) . (1.49)

In momentum space (1.48) simply reads

[∂2

∂t2+(p2 +m2

)]φ(t,p) = 0 , (1.50)

which tells us that for each value of p, the Fourier transform φ(t,p) solves the equation of aharmonic oscillator with frequency

ωp =√|p|2 +m2 . (1.51)

It is then not difficult to see that the most general solution of the classical Klein-Gordonequation is a linear superposition of simple harmonic oscillators,5

φ(x) =

∫d3p

(2π)31

2ωp

(a(p)e−ipx + a∗(p)eipx

), (1.52)

each vibrating at a different frequency with a different amplitude (the appearance of thespecific combination of coefficients a(p) and a∗(p) is dictated by the reality of the Klein-Gordon field φ). The 4-vector p in the exponents is understood as pµ = (wp,p). Whilethe decomposition (1.52) has not actually been derived here, the result looks plausible afternoticing that the measure d3p/(2π)3 1/(2ωp) is a Lorentz-invariant quantity. This follows from“reverse engineering”

∫d3p

(2π)31

2ωp=

∫d3p

(2π)31

2p0

∣∣∣∣p0=ωp

=

∫d4p

(2π)3δ(p20 − p2 −m2)

∣∣p0>0

. (1.53)

5The actual derivation of this result can be found on page 47 of the script by John Chalker and Andre Lukas.It is not repeated here.

11

From the latter result we can also figure out that the Lorentz-invariant delta function for3-momenta is

2ωp δ(3)(p− q) , (1.54)

since ∫d3p

2ωp2ωp δ

(3)(p− q) = 1 . (1.55)

The Lorentz invariance of the measure d3p/(2π)3 1/(2ωp) and its consequences, should be keptin mind, because these features will be important at several occasions later on.

Let me add that in order to quantize the field φ, we must hence only quantize a infi-nite number of harmonic oscillators (as Sidney Coleman once said: “The career of a youngtheoretical physicist consists of treating the harmonic oscillator in ever-increasing levels ofabstraction.”). You should already know how this is done in QM and we will see later on howit is done in QFT.

Lorentz Invariance

Let us first check that a LTφ(x)→ φ′(x) = φ(Λ−1x) , (1.56)

leaves the the KG action invariant. Notice that the inverse Λ−1 appears here because we aredealing with an active transformation, in which the field is truly shifted. To see why thismeans that the inverse appears, it will suffice to consider a non-relativistic example such asa temperature field. Suppose we start with an initial field φ(x) which has a hotspot at, say,x = (1, 0, 0). Let’s now make a rotation x → Ox about the z-axis so that the hotspot endsup at x = (0, 1, 0). If we want to express the new field φ′(x) in terms of the old field φ(x),we have to place ourselves at x = (0, 1, 0) and ask what the old field looked like at the pointO−1x = (1, 0, 0) we came from. This O−1 is the origin of the Λ−1 factor in the argument ofthe transformed field in (1.56).

According to the LT (1.56), the transformation of the mass term is 1/2m2φ2(x) →1/2m2φ2(x′) with x′ = Λ−1x. The transformation of ∂µφ is

∂µφ(x)→ ∂µ(φ(x′)) = (Λ−1)νµ(∂νφ)(x′) . (1.57)

Here we have used that

∂µ =∂

∂xµ=

∂

∂(x′)ν∂(x′)ν

∂xµ= (Λ−1)νµ ∂

′ν . (1.58)

and wrote ∂′µφ(x′) = (∂µφ)(x′). Below we will also use the notation ∂µφ(x′) to denote thederivative at x = x′. Using the defining property (Λ−1)ρµ(Λ−1)σν η

µν = ηρσ of the LTs, wethus find that the derivative term in the Klein-Gordon Lagrangian behaves as

1

2(∂µφ(x))2 → 1

2(Λ−1)ρµ(∂ρφ)(x′)(Λ−1)σν(∂σφ)(x′) ηµν

=1

2(∂ρφ)(x′)(∂σφ)(x′) ηρσ

=1

2(∂µφ(x′))

2.

(1.59)

12

Putting things together, we see that the action of the Klein-Gordon theory is indeed Lorentzinvariant,

S =

∫d4xL(x)→

∫d4xL(x′) =

∫d4x′ L(x′) = S . (1.60)

Notice that changing the integration variables from d4x to d4x′, in principle introduces anJacobian factor det (Λ). This factor is, however, equal to 1 for LT connected to the identity,that we are dealing with (remember that 14 is part of the proper, orthochronous or restrictedLorentz group L↑+).

A similar calculation also shows that, as promised, also EOM of the Klein-Gordon field φ,as given in (1.47), is invariant,

(∂2 +m2

)φ(x)→

(∂2 +m2

)φ′(x)

=[(Λ−1)νµ∂ν(Λ

−1)ρµ∂ρ +m2]φ(x′)

=(ηνρ∂ν∂ρ +m2

)φ(x′) = 0 .

(1.61)

Here we have used another common notation for the Laplacian, ∂µ∂µ = ∂2.Notice that the above explicit example shows that the Lagrangian formulation of field

theory makes it especially easy to discuss Lorentz invariance, since an EOM is automaticallyLorentz invariant if it follows from a Lagrangian that is a Lorentz scalar. This is an immediateconsequence of the principle of least action. If a LT leaves the Lagrangian unchanged, thetransformation of an extremum in the action will be another extremum.

Conservation Laws

Let us derive the Hamiltonian density and the energy-momentum tensor for (1.42). Theconjugate momentum of φ is simply π = φ, and therefore

H =1

2π2 +

1

2(∇φ)2 + V . (1.62)

For the stress tensor we find from (1.37) the expression

T µν = (∂µφ)(∂νφ)− 1

2ηµν (∂ρφ)2 + ηµν V . (1.63)

Notice that the energy-momentum tensor of (1.42) immediately comes out symmetric, i.e.,Tµν = Tνµ. This is not always the case. In accordance with our general formula (1.40), wehence have

P 0 =

∫d3xT 00 =

∫d3x

(1

2π2 +

1

2(∇φ)2 + V

)=

∫d3xH = H . (1.64)

The energy is hence conserved in our real scalar theory. Analog relations hold for the spacecomponents P i, signalling 3-momentum conservation.

In classical particle mechanics, rotational invariance gives rise to conservation of angularmomentum. What is the analogy in field theory? What happens to the remaining three LTs,

13

namely the boosts. What conserved quantity do they correspond to? In order to address thesequestions, we first need the infinitesimal form of the LTs

Λµν = δµν + ωµν , (1.65)

where ωµν is infinitesimal. The condition (1.2) for Λ to be a LT becomes in infinitesimal form

ηµν = ηρσ (δµρ + ωµρ) (δνσ + ωνσ) = ηµν + ωµν + ωνµ +O(ω2) . (1.66)

This shows that ωµν must be an anti-symmetric matrix (a feature that I have stated beforewithout proof),

ωµν = −ωνµ . (1.67)

Notice that an anti-symmetric 4×4 matrix has six independent parameters, which agrees withthe number of different Lorentz transformations, i.e., three rotations and three boosts.

Applying the infinitesimal LT to our real scalar field φ, we have

φ(x)→ φ(x− ωx) = φ(x)− ωµνxν∂µφ(x) , (1.68)

where the minus sign arises from the factor Λ−1 in (1.56). The variation of the field φ underan infinitesimal LT is hence given by

δφ = −ωµνxν∂µφ . (1.69)

By the same line of reasoning, one shows that the variation of the Lagrangian is

δL = −ωµνxν∂µL = −∂µ (ωµνxνL) , (1.70)

where in the last step we used the fact that ωµµ = 0 due to its anti-symmetry. The Lagrangianchanges by a total derivative, so we can apply Noether’s theorem (1.34) with J µ = −ωµνxνLto find the conserved current,

Jµ = − ∂L∂(∂µφ)

ωρνxν∂ρφ+ ωµνx

νL

= −ωρν[

∂L∂(∂µφ)

∂ρφ− δµρL]xν = −ωρνT µρxν .

(1.71)

Stripping off ωρν , we obtain six different currents, which we write as

(Iλ)µν = xµT λν − xνT λµ . (1.72)

These currents satisfy∂λ(Iλ)µν = 0 , (1.73)

and give (as usual) rise to six conserved charges . For µ, ν 6= 0, the LT is a rotation andthe three conserved charges give the total angular momentum of the field configuration (i, j =1, 2, 3):

Qij =

∫d3x

(xiT 0j − xjT 0i

). (1.74)

14

What’s about the boosts? In this case, the conserved charges are

Q0i =

∫d3x

(x0T 0i − xiT 00

). (1.75)

The fact that these are conserved tells us that

0 =dQ0i

dt=

∫d3xT 0i + t

∫d3x

dT 0i

dt− d

dt

∫d3x xiT 00

= P i + tdP i

dt− d

dt

∫d3x xiT 00 .

(1.76)

Yet, also the momentum P i is conserved, i.e., dP i/dt = 0, and we conclude that

d

dt

∫d3x xiT 00 =

d

dt

∫d3x xiH = const. (1.77)

This is the statement that the center of energy of the field travels with a constant velocity. Ina sense it’s a field theoretical version of Newton’s first law but, rather surprisingly, appearinghere as a conservation law. Notice that after restoring the label a our results for (Iλ)µν etc.also apply in the case of multicomponent fields.

Poincare Invariance

We now require that a physical system possesses both space-time translation (1.35) and Lorentzsymmetry. The symmetry group that includes both transformations is called the Poincaregroup. Notice that for any Poincare-invariant theory the two charge conservation equations(1.38) and (1.73) should hold. This is only possible if the energy-momentum tensor T µν issymmetric. Indeed,

0 = ∂λ(Iλ)µν = ∂λ(xµT λν − xνT λµ

)

= xµ∂λTλν + T λν∂λx

µ − xν∂λT λµ − T λµ∂λxν

= T λνδλµ − T λµδλν = T µν − T νµ .

(1.78)

This general result tells us that the expression of the energy-momentum tensor in any Poincare-invariant theory can be made symmetric without changing physics. The key to actually do it,lies in making use of the conservation law (1.38) in an appropriate way.

Discrete Internal Symmetries

So far we have only imposed external symmetries, such as Lorentz or Poincare symmetry, onour scalar theory. Let us also have a brief look at internal symmetries. An interesting internalsymmetry to be considered is a Z2 symmetry which acts as

φ(x)→ −φ(x) . (1.79)

15

-v +vΦ

VHvL

VHΦL

Figure 1.3: Shape of the scalar potential for m2 ≥ 0 (solid line) and m2 < 0 (dashedline). In the latter case the positions of the minima are at φ = ±v = ±

√(−6m2)/λ.

The value of the potential at the minima is V (v) = V0 − (λv4)/24.

This transformation leaves (1.42) invariant if the scalar potential V contains only terms withan even number of φ fields (but it does not lead to a conserved quantity since it is not acontinuous symmetry). Restricting ourselves to relevant and marginal couplings, we write

V = V0 +1

2m2φ2 +

λ

4!φ4 . (1.80)

where we have added a constant term V0 to the potential. Notice that such a shift leaves theEOM (1.46) unaltered.

Let us now study the simplest type of solutions to the real scalar theory with the poten-tial (1.80). A trivial solution of (1.46) is φ(x) = v = const., provided that

∂V

∂φ

∣∣∣∣φ=v

= 0 . (1.81)

This relation implies that we should look for the extrema of the potential (1.80). In fact,to minimize the energy (1.64) we should focus on the minima of V . Such solutions of theclassic theory are called vacua (or vacuum if there is a single one). Note that if the quarticcoupling λ is negative, the potential is not bounded from below and the energy of a constantfield configuration tends to minus infinity for large field values. To avoid such an unphysicalsituation, we simply assume λ > 0 in what follows.

The shape of the scalar potential V is shown in Figure 1.3. We see that one has todistinguish two cases. First, the case where m2 ≥ 0 (solid curve). In this situation there isa single minimum located at the origin φ = 0. This solution is mapped into itself under thetransformation (1.79) and we say that the symmetry is unbroken in this vacuum. The second

16

case with m2 < 0 (dashed curve) is more interesting. In this case φ = 0 is a maximum, whilethere are two minima. These are located at

φ = ±v = ±√−6m2

λ. (1.82)

At the minima, the potential takes the value

V (v) = V0 +1

4m2v2 = V0 −

λv4

24. (1.83)

Notice that neither minimum is invariant under the Z2 symmetry. In fact, the minima areinterchanged under (1.79), and we say that the symmetry is spontaneously broken. In general,spontaneous symmetry breaking refers to the case where a symmetry of the theory is partiallyor fully broken by the vacuum solution of the theory. We will see later on that the concept ofspontaneous symmetry breaking is a rather important one.

Continuous Internal Symmetries

Besides discrete also continuous internal symmetries play an important role in particle physics.We start our discussion by considering a scalar theory that involves two real scalar fields φ1,2

and combine these fields into a field vector

φ =

(φ1

φ2

). (1.84)

In the following we will be interested in theories that are invariant under SO(2) transformations(aka 2-dimensional rotations) that take the form

φ→ φ′ = Oφ . (1.85)

Since this transformation is the same at every space-time point, symmetries of this type arecalled global. The length of φ should be SO(2) invariant, which implies that

OTO = OOT = 12 , (1.86)

and we also require thatdet (O) = +1 . (1.87)

Proper 2-dimensional rotation matrices can hence be written as

O = eiαT , (1.88)

where α is a real parameter (the rotation angle) and T denotes the 2×2 matrix that generatesthe 2-dimensional rotations. A suitable choice of the generator reads

T =

(0 −ii 0

). (1.89)

17

Notice that SO(2) is said to be an abelian group, since

eiαT eiβT = eiβT eiαT = ei(α+β)T , (1.90)

which means that the ordering of two successive group transformations does not matter.In the case of 3-dimensional rotations things become slightly more complicated. Since

there is an independent rotation corresponding to each plane in three dimensions, the SO(3)group has 3 (3− 1) /2 = 3 parameters αi (Euler angles) and three generators Ti (i = 1, 2, 3).Infinitesimal SO(3) transformations thus act on φ = (φ1, φ2, φ3)

T as φ → φ′ = φ + iαiTiφ.From (1.86) and (1.87) it follows that the generator Ti are hermitian and traceless:

(Ti)† = Ti , Tr (Ti) = 0 . (1.91)

A possible set of 3× 3 matrices with these properties is

T1 =

0 0 0

0 0 −i0 i 0

, T2 =

0 0 i

0 0 0

−i 0 0

, T3 =

0 −i 0

i 0 0

0 0 0

, (1.92)

which can be written in a more compact way as (Ti)jk = −iεijk (j, k = 1, 2, 3). By a straight-forward calculation, one can prove that the latter matrices satisfy the following Lie algebra

[Ti, Tj] = iεijkTk . (1.93)

These relations imply that two 3-dimensional rotations do in general not “commute”, whichcauses the non-abelian character of SO(3).

Other examples of non-abelian groups are provided by the special unitary groups SU(n),which consist of all complex n× n matrices U satisfying

U †U = UU † = 1n , det (U) = +1 . (1.94)

The simplest non-trivial example of a special unitary group is SU(2). It has 22 − 1 = 3generators τi and close to the identity one can write the group transformations as U = 12+itiτi.From (1.94) it follows that the generators obey τ †i = τi and Tr (τi) = 0. A convenient choiceof the generators τi is

τi =σi2, (1.95)

where σi are the usual Pauli matrices given by

σ1 =

(0 1

1 0

), σ2 =

(0 −ii 0

), σ3 =

(1 0

0 −1

). (1.96)

Recalling that σiσj = δij12 + iεijkσk, it is easy to show that Tr (τiτj) = δij/2 and that theSU(2) generators fulfil the commutation relations

[τi, τj] = iεijkτk . (1.97)

18

Interestingly, (1.93) and (1.97) are the same commutation relations. This means thatthe matrices τi and Ti form two different representations of the SU(2) Lie algebra. Therepresentation that is classified by the 2 × 2 matrices (1.95) is called fundamental. It actson a complex vector of size two and hence the dimension of the representation space is two.The representation where the generators are 3 × 3 matrices of the form (Ti)jk = −iεijk

(as

in (1.92))

is instead called the adjoint representation of SU(2). This representation acts on3-dimensional real vectors and therefore it has the dimension of SU(2) that is equal to three.

A Complex Scalar Field With a U(1) Symmetry

For the further discussion it will prove convenient to arrange the two real scalar fields thatwe have meet at the beginning of the previous subsection into a single complex Klein-Gordonfield defined as

ϕ =1√2

(φ1 + iφ2) . (1.98)

On this complex representation, the transformation (1.85) acts as follows

(φ1

φ2

)→(

cosα − sinα

sinα cosα

)(φ1

φ2

)=⇒

(φ1

iφ2

)→(

cosα i sinα

i sinα cosα

)(φ1

iφ2

)

=⇒ φ1 + iφ2 → eiα (φ1 + iφ2) .

(1.99)

This means that all terms in the Lagrangians we are looking for should be invariant under afield phase-redefinition aka a global U(1) transformation,

ϕ→ ϕ′ = eiαϕ , ϕ∗ → (ϕ′)∗ = e−iαϕ∗ . (1.100)

The different transformation properties of ϕ and ϕ∗ tell us that these fields carry charge 1 and−1, respectively.

Since ϕ and ϕ∗ carry a non-zero charge not all terms polynomial in the fields are allowedin the Lagrangian if we want to respect the symmetry. E.g., a term ϕ2 has charge +2 and ishence not invariant under (1.100). In general, we can only allow terms with the same numberof ϕ and ϕ∗ field, so that the sought U(1) invariant Lagrangian takes the form

L = (∂µϕ∗)(∂µϕ)− V (ϕ∗ϕ) , (1.101)

with

V (ϕ∗ϕ) = V0 +m2ϕ∗ϕ+λ

4(ϕ∗ϕ)2 . (1.102)

Note that it is essential for the invariance of the kinetic term that the group parameter α isspace-time independent. For the EOM for ϕ, we find from the Euler-Lagrange equation (1.22),

∂2ϕ+∂V

∂ϕ∗= ∂2ϕ+m2ϕ+

λ

2(ϕ∗ϕ)ϕ = 0 . (1.103)

19

A similar equation holds in the case of ϕ∗. The conjugate momenta are π = ϕ∗ and π∗ = ϕ,so that the Hamiltonian reads

H = πϕ+ π∗ϕ∗ − L = π∗π + (∇ϕ∗) · (∇ϕ) + V (ϕ∗ϕ) . (1.104)

Being translation and Lorentz invariant the above theory has conserved stress and angularmomentum tensors which can be obtained in complete analogy with the single scalar fieldcase discussed before. We leave the derivation of the explicit expressions for T µν and Qij asan useful exercise.

According to Noether’s theorem the presence of the internal symmetry (1.100) leads to anew type of conserved current which we will now derive. The variations of the fields ϕ and ϕ∗

(treated as independent) under the global U(1) symmetry are

δϕ = iαϕ , δϕ∗ = −iαϕ∗ . (1.105)

By plaguing these variations into the general formula (1.34) and realizing that in our caseJ µ = 0 since δL = 0, the conserved current and charge is readily written down:

Jµ = i (ϕ∗ ∂µϕ− ϕ∂µϕ∗) , Q = i

∫d3x (ϕ∗ ϕ− ϕϕ∗) = i

∫d3x (ϕ∗π∗ − ϕπ) . (1.106)

There is an ambiguity worth noting, when applying Noether’s theorem to find the conservedcharge under the transformation (1.100). Obviously, if Q is conserved, then so is every otheroperator c1Q + c2 with c1,2 constant numbers. The expression for Q in (1.106) is thereforeunique up to a multiplicative and an additive constant. The multiplicative factor essentiallydenotes the units in which we measure the charge. Notice that we have already used this am-biguity in (1.106) and simply ignored a factor −α. Later we will also learn how the ambiguityon the additive constant is removed.

Spontaneous Symmetry Breaking

We would now like to discuss the vacua of the theory, i.e., solutions to the EOM (1.103) withϕ = v = const. Like in the real scalar case for m2 ≥ 0 there is a single minimum at ϕ = 0.This solution is left invariant under the transformations (1.100) and hence the U(1) symmetryis unbroken. For m2 < 0 the shape of the potential is illustrated in Figure 1.4. We can seefrom the depicted red line that in this case there is a whole circle of minima

ϕ = v =v0√

2eiδ , v0 =

√−4m2

λ, (1.107)

where δ is an arbitrary phase. Clearly, the existence of this one-dimensional space of vacua isnot an accident, but originates from the U(1) invariance of the scalar potential (1.102):

V (ϕ∗ϕ)→ V ((ϕ′)∗ϕ′) = V (e−iαϕ∗eiαϕ) = V (ϕ∗ϕ) . (1.108)

In fact, this invariance implies that for every minimum ϕ of V also ϕ′ = exp (iα)ϕ is aminimum for arbitrary α. We can use this freedom to choose a particular minimum that

20

Φ1

Φ2

V

Φ1

Φ2

V

ϕ1

ϕ2

Figure 1.4: Left: Shape of the scalar potential of the complex Klein-Gordon fieldfor m2 < 0. The positions of the minima are indicated by the red curve. Right:Field excitations ϕ1,2 (black arrows) around the minimum of the scalar potential atϕ = v0

√2.

breaks the U(1) symmetry spontaneously. Let’s take the minimum on the φ1 axis, meaningϕ = v0/

√2. Around this point we then can expand the field as

ϕ =1√2

(v0 + ϕ1 + iϕ2) , (1.109)

where ϕ1,2 are understood to be small fluctuations. Inserting the above relation into (1.102),we find

V = V0 +1

4m2v20 −m2ϕ2

1 +O(ϕ31, ϕ1ϕ

22) . (1.110)

This is an interesting result, because it tells us that while the excitation ϕ1 is massive withmass

√2m (remember m2 < 0) the excitation ϕ2 remains massless. Looking at Figure 1.4 this

is not a completely unexpected feature. While ϕ2 corresponds to an fluctuation around thecircle of minima (a flat direction in the potential), ϕ1 represents an excitation perpendicularto ϕ1 (a direction of the potential with a curvature). The appearance of massless scalars forspontaneously broken global symmetries is a general feature known as Goldstone’s theoremand the corresponding massless scalars are also called Goldstone bosons. We will now studythis phenomenon in a more general setting.

Goldstone’s Theorem

To generalize the above findings, we consider a field theory involving a set of scalar fieldsφ = (φ1, . . . , φn)T with corresponding potential V (φ). We assume that the potential has aminimum at v = (v1, . . . , vn)T that satisfies

∂V

∂φa

∣∣∣∣φ=v

= 0 . (1.111)

21

After defining ϕ = φ− v, we can Taylor expand the potential around such a minimum in thefollowing way

V = V (v) +1

2Mabϕ

aϕb +O(ϕ3) , (1.112)

where (a, b = 1, . . . , n)

Mab =∂2V

∂φa∂φb

∣∣∣∣φ=v

. (1.113)

Clearly, the eigenvalues of the mass matrix M are the mass squares of the fields (around thevacuum v).

Now let us assume that our scalar theory is invariant under a continuous (global) symmetrygroup G, under which

φ→ R(g)φ , (1.114)

where R(g) denotes the representation of φ. The invariance of the scalar theory in particularmeans that the scalar potential is invariant under such a transformation, namely

V (φ) = V (R−1(g)φ) , (1.115)

for all g ∈ G. The vacuum will in general not respect the full symmetry group G, but willspontaneously break it into a subgroup H ⊂ G, so that

R(g)v = v for g ∈ H , R(g)v 6= v for g 6∈ H . (1.116)

Now introducing infinitesimal group transformations

R(g) = 1 + itαTα , (1.117)

with generators Tα (in the representation R) and small coefficients tα. The set of generators{Tα} can be split into two subsets Ti and Tj, where Ti denotes the generators of the unbrokensubgroup H, while Tj are the remaining generators corresponding to the broken part of G. Bydefinition, one has

Tiv = 0 , Tjv 6= 0 . (1.118)

Now let us see how (1.117) acts on our potential. We have

V (φ) = V (φ− itαTαφ) = V (φ)− i(∂V

∂φ

)TtαTαφ . (1.119)

Differentiating this expression again and evaluating it at φ = v, one finds using (1.111)and (1.113), that

MTαv = 0 . (1.120)

From this relation we see that every broken generator Tj (with Tjv 6= 0) leads necessarily toan eigenvector of M with eigenvalue zero. Or in more physical terms, every broken generatorleads to a massless scalar excitation. We have proven Goldstone’s theorem!

22

Scalar Field Theory With a SU(2)× U(1) Symmetry

Let me further illustrate Goldstone’s theorem by a still simple but educated example. Weconsider the symmetry SU(2) × U(1) and a complex scalar field Φ that is a doublet underSU(2) and has a charge of 1/2 under the U(1). Our scalar transforms as

Φ→ eiαY eitiτi Φ '

(1 + iαY + itiτi

)Φ , (1.121)

with the generators

Y =1

2, τi =

σi2. (1.122)

Notice that our notation is on purpose sloppy here, since the 1 on the right-hand side of (1.121)should be in fact 12. A similar statement also applies to the generator Y . Such a sloppinessin notation is common in textbooks, and it is good to get used to it early.

The Lagrangian we are interested in, can be written as

L = (∂µΦ)† (∂µΦ)− V (Φ†Φ) (1.123)

withV (Φ†Φ) = V0 + µ2 Φ†Φ + λ

(Φ†Φ

)2. (1.124)

Provided that λ > 0 and µ2 < 0 (again a change in notation), the scalar potential (1.124) isbounded from below and minimized for

Φ†Φ = v20 = −µ2

2λ. (1.125)

Using (1.121) allows us to choose a particular simple vacuum configuration. We take

Φ = v =

(0

v0

). (1.126)

For this choice, one finds readily

τ1v 6= 0 , τ2v 6= 0 , (τ3 − Y )v 6= 0 , (1.127)

and(τ3 + Y )v = 0 . (1.128)

Hence three of the four generators of SU(2) × U(1) are broken, which corresponds to thebreaking pattern

SU(2)× U(1)→ U(1) , (1.129)

where the U(1) on the right-hand side is the diagonal subgroup of SU(2)×U(1) correspondingto the generator Q = τ3 + Y .

So why is this example relevant? The relevance of the example stems from the fact thatthe breaking pattern (1.129) is precisely that of the electroweak gauge group of the SM ofparticle physics: the SM is in fact based on SU(3)c × SU(2)L × U(1)Y → SU(3)c × U(1)em,

23

where the subscript c refers to color, L indicates that the SU(2) only acts on left-handedfields, Y represents hypercharge, and U(1)em corresponds to the unbroken electromagnetismassociated to the generator Q, i.e., the electric charge. Yet, there is an important difference.While our theory predicts three massless Goldstone bosons, in Nature we only observe a singleweakly-interacting massless state, i.e., the photon. This difference is explained by the factthat the electroweak symmetry in the SM is realized as a local or gauge(d) symmetry, whilein our simple example we dealt with a global symmetry. It turns out that in the case of thespontaneously broken electroweak gauge symmetry our three massless Goldstone bosons areabsorbed (or eaten) by three vector bosons (the electrical neutral Z boson and the chargedW± bosons) that receive their mass from the breaking mechanism. This is the famous Higgsmechanism, which also gives rise to a (physical) massive scalar state the Higgs (sometimesalso referred to as the “god particle”). Let me mention that in 2013 both the ATLAS and theCMS collaborations (situated at the CERN LHC in Geneva) have announced the existenceof a new bosonic state with a mass of around 125 GeV. This discovery defines a turningpoint in the history of elementary-particle physics: the almost 50 year-long hunt for theHiggs boson has come to a dazzling end. This year the Nobel committee gave the Nobelprize to Francois Englert and Peter W. Higgs for their theoretical work on the mechanismof electroweak symmetry breaking. In order to fully understand the physics of the Higgsmechanism, we are still missing an important ingredient, namely vector fields. These 4-component fields will be discussed now.

1.5 Vector Field Theories

In this section we will examine the basic features of another important ingredient of realistic(quantum) field theories, i.e., vector fields. These fields carry spin one and are called vectorfields, since they transform like a vector under the Lorentz group.

Maxwell’s Theory Of Electromagnetism

As another simple application of the formalism we have developed, let us try to deriveMaxwell’s equations using the field theory formulation. In terms of the electric and mag-netic fields E and B and the charge density ρ and 3-vector current j, these equations takethe well-known form

∇ ·B = 0 , (1.130)

∇×E +∂B

∂t= 0 , (1.131)

∇ ·E = ρ , (1.132)

∇×B − ∂E

∂t= j . (1.133)

The E and B fields are spatial 3-vectors and can be expressed in terms of the components

24

of the 4-vector field Aµ = (φ,A) by

E = −∇φ− ∂A

∂t, B =∇×A . (1.134)

This definition ensures that the first two homogeneous Maxwell equations (1.130) and (1.131)are automatically satisfied,

∇ · (∇×A) = εijk∂i∂jAk = 0 , (1.135)

∇×(−∇φ− ∂A

∂t

)+∂

∂t(∇×A) = −∇× (∇φ) = −εijk∂j∂kφ = 0 . (1.136)

Here the indices i, j, k = 1, 2, 3 are summed over if they appear twice.The remaining two inhomogeneous Maxwell equations (1.132) and (1.133) follow from the

Lagrangian

L = −1

2(∂µAν) (∂µAν) +

1

2(∂µAν) (∂νAµ)− AµJµ , (1.137)

with Jµ = (ρ, j). From the rules presented before, we gather that the dimension of the vectorfield and current is [Aµ] = 1 and [Jµ] = 3, respectively. The funny minus sign of the first termon the right-hand side is required to ensure that the kinetic term 1/2A2

i is positive using theMinkowski metric. Notice also that the Lagrangian (1.137) has no kinetic term 1/2A2

0 andhence A0 is not dynamical. Why this is and necessarily has to be the case will become clearpretty soon.

Enough said, let’s do serious business and compute something. To see that the statementmade before (1.137) is indeed correct, we first evaluate

∂L∂Aν

= −Jν , ∂L∂(∂µAν)

= −∂µAν + ∂νAµ , (1.138)

from which we derive the EOMs,

0 = ∂µ

(∂L

∂(∂µAν)

)− ∂L∂Aν

= ∂µ(−∂µAν + ∂νAµ) + Jν . (1.139)

Introducing now the field-strength tensor

Fµν = ∂µAν − ∂νAµ , (1.140)

we can write (1.137) and (1.139) quite compact,

L = −1

4FµνF

µν − JµAµ . (1.141)

∂µFµν = Jν , (1.142)

Does this look familiar? I hope so. Notice that we have [Fµν ] = 2 and that the field-strengthtensor satisfies the Bianchi identity:

∂ρFµν + ∂µFνρ + ∂νFρµ = 0 . (1.143)

25

In order to see that (1.142) indeed captures the physics of (1.132) and (1.133), we computethe components of F µν . We find

F 0i = −F i0 = ∂0Ai − ∂iA0 =

(∇φ+

∂A

∂t

)i= −Ei ,

F ij = −F ji = ∂iAj − ∂jAi = −εijkBk ,

(1.144)

while all other components are zero. With this in hand, we then obtain from ∂µFµ0 = ρ and

∂µFµ1 = j1,

∂µFµ0 = ∂0F

00 + ∂iFi0 =∇ ·E = ρ ,

∂µFµ1 = ∂0F

01 + ∂iFi1 = −∂E

1

∂t+∂B3

∂x2− ∂B2

∂x3=

(∇×B − ∂E

∂t

)1

= j1 .(1.145)

Similar relations hold for the remaining components i = 2, 3. Taken together this proves thesecond inhomogeneous Maxwell equation (1.133).

Let me also derive the energy-momentum tensor T µν of electrodynamics, ignoring for themoment the source term JµA

µ. Using (1.138) one finds

T µν = −F µρ∂νAρ +1

4ηµνFρσF

ρσ . (1.146)

Notice that the first term in (1.146) is not symmetric, which implies that T µν 6= T νµ. In fact,this is not really surprising since the definition of the energy-momentum tensor (1.37) doesnot exhibit an explicit symmetry in the indices µ and ν. Nevertheless, there is always a wayto massage the energy-momentum tensor of any theory into a symmetric form.6 To learn howthis can be done in the case under consideration is the objective of a homework problem.

Physical Degrees Of Freedom

Under LTs, Aµ transforms as a vector, i.e., Aµ → ΛµνAν . The same applies of course also

to the current Jµ. The field-strength tensor Fµν , on the other hand, carries two indices andhence transforms as a tensor, i.e., Fµν → Λµ

ρΛνσFρσ. Equipped with these transformation

properties it is an easy exercise to show that the Lagrangian (1.141) is Lorentz invariant (atthe end it is sufficient to observe that in the expression for L all indices are contracted, so that(1.141) is a Lorentz scalar). The explicit calculation is left as an exercise to the interestedreader.

After (1.137) we have already mentioned that not all components of Aµ are dynamical,which signals that Aµ contains unphysical dofs. In fact, this is not really a surprise because itis Fµν that is directly related to the physical fields E and B not Aµ (see (1.144)). Formally,this is expressed by the fact that a gauge transformation

Aµ(x)→ Aµ(x) + ∂µf(x) , (1.147)

6One (but not the only) reason that you might want to have a symmetric energy-momentum tensor Tµν

is to make contact with general relativity, since such an object sits on the right-hand side of Einstein’s fieldequations.

26

with arbitrary function f(x) leaves Fµν invariant:

Fµν = ∂µAν − ∂νAµ → ∂µ(Aν + ∂νf)− ∂ν(Aµ + ∂µf)

= ∂µAν − ∂νAµ + ∂µ∂νf − ∂ν∂µf = Fµν .(1.148)

This is an important result, because it gives us a recipe on how to “derive” the covariantformulation of electrodynamics without ever talking about Maxwell’s equations. The idea issimply to find the most general Lagrangian (including terms up to a certain mass dimension)that is Lorentz invariant and unchanged under (1.147). Notice that gauge invariance in partic-ular implies that L should only depend on Aµ through Fµν . Up to dimension four this leavesbasically only one term7 (in the case of a vanishing source term), namely L = −1/4FµνF

µν .The associated EOM is of course ∂µF

µν = 0, or written in terms of Aµ,

∂2Aµ − ∂µ∂νAν = 0 . (1.149)

In fact, this equation can be further simplified by exploiting the gauge symmetry (1.147) byimposing a gauge condition on Aµ through a suitable choice of gauge parameter f . There arevarious possibilities for such a gauge choice and here we consider the Lorenz gauge defined by

∂µAµ = 0 . (1.150)

This gauge has the salient benefit that it is covariant (in contrast to other gauges such astemporal gauge A0 = 0 or Coloumb gauge ∇ ·A = 0) and it simplifies (1.149) to

∂2Aµ = 0 . (1.151)

It is however important to realize that (1.150) does not fix the gauge entirely but leaves aresidual gauge freedom, since all choices of f with

∂2f = 0 , (1.152)

are equivalent in the sense that they all lead to (1.151).The solutions of ∂2Aµ = 0 are immediately written done after noticing that the latter

equation is nothing but a Klein-Gordon equation for a massless vector field. Adding vectorindices to (1.52), we arrive at

Aµ(x) =

∫d3p

(2π)31

2ωp

(aµ(p)e−ipx + a∗µ(p)eipx

), (1.153)

where ωp = |p| and pµ = (ωp,p). Fourier transforming the Lorenz gauge condition (1.150),we find in addition

pµaµ(p) = 0 . (1.154)

7The term Fµν Fµν with Fµν = εµνρσFρσ denoting the dual field-strength tensor also mets all requirements.

It can however be written as a total derivative 4∂µ(εµνρσAν∂ρAσ) and therefore does not contribute to theclassic EOMs. We hence ignore such a term.

27

To exploit this constraint one conveniently introduces a set of polarization vectors ε(α)µ (p)

with α = 0, 1, 2, 3 and the following properties. The vectors ε(1)µ and ε

(2)µ are orthogonal to

both p and a vector n with n2 = 1 and n0 > 0. They furthermore are chosen such that theysatisfy ε(α) · ε(β) = −δαβ for α, β = 1, 2. The polarization vector ε

(3)µ is taken to be in the p–n

plane, orthogonal to n and normalized to -1, i.e., n · ε(3) = 0 and (ε(3))2 = −1. Finally, onedefines ε(0) = n. With these conventions one has an orthogonal set of vectors satisfying

ε(α) · ε(β) = ηαβ , (1.155)

for α, β = 0, 1, 2, 3. This basis can be used to write

aµ(p) =3∑

α=0

a(α)(p)ε(α)µ (p) . (1.156)

In fact, the basic idea behind all this “mumbo-jumbo” is that this specific choice of basis ofpolarization vectors, allows to easily separate the directions transversal to p (corresponding to

ε(1)µ and ε

(2)µ ) from the other two directions (corresponding to ε

(0)µ and ε

(3)µ ) that are longitudinal

to p. As an example, if we choose a spatial momentum p pointing in the z-directions, thecomponents of above vectors are explicitly given by

ε(α)µ = δαµ . (1.157)

In the general case, one has instead

ε(3)µ =pµp0− nµ . (1.158)

Clearly,

n · ε(3) =np

p0− n2 =

np

p0− 1 = 0 =⇒ np

p0= 1 , (1.159)

and thus

(ε(3))2 =p2

p20+ n2 − 2

np

p0= 1− 2

np

p0= −1 , (1.160)

where we have used that our vector field is massless, i.e., p2 = 0. So (1.158) is indeed thesought solution. Notice also that

ε(0)µ + ε(3)µ =pµp0. (1.161)

We now return to (1.154) and evaluate this sum inserting the expansion (1.156). By

definition the transversal components ε(1)µ and ε

(2)µ do not contribute and we find

pµaµ(p) = (p · ε(0))a(0)(p) + (p · ε(3))a(3)(p) = p0(a(0)(p)− a(3)(p)

)= 0 . (1.162)

Here we have used thatp · ε(0) = −p · ε(3) = p0 . (1.163)

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Note that the first equality follows from (1.161) after contracting it with pµ, while the secondequality is a consequence of (1.159). From (1.162) it follows that a(0)(p) = a(3)(p). Theexpansion (1.156) can thus be written as

aµ(p) = a(3)(p)pµp0

+2∑

α=1

a(α)(p)ε(α)µ (p) , (1.164)

and we are left with two transversal modes and a longitudinal one along the p direction(remember that after (1.137) we already mentioned the fact that the time component A0 isnot dynamical, which reduces the number of independent dofs in Aµ from four to three). Thisis still one more dof than a physical massless field, such as the photon, ought to have.

The trick to get rid of the remaining longitudinal component is to make use of the residualgauge freedom (1.152) and to “gauge away” the term proportional to a(3)(p) in (1.164). Letsee how this works. Given that f also fulfils a Klein-Gordon equation, the most generaldecomposition obviously reads

f(x) =

∫d3p

(2π)31

2ωp

(ξ(p)e−ipx + ξ∗(p)eipx

). (1.165)

A gauge transformation (1.147) with such a f corresponds to

aµ(p)→ a′µ(p) = aµ(p)− ipµξ(p) . (1.166)

But such a shift is exactly what is needed to remove the remaining longitudinal componentfrom (1.164) and we arrive at the correct physical answer. Needless to say that the reduc-tion from apparent four dofs to actually two is intimately related to the gauge invariance of(quantum) electrodynamics.

A Massive Vector Field

What happens if we try to give our vector field Aµ a mass? We add a term

1

2m2AµA

µ , (1.167)

to our Lagrangian L = −1/4FµνFµν , but immediately realize that such a term is not allowed

by gauge invariance (1.147). This is of course stupid, but let us ignore this unwanted featurefor a moment. The EOMs for the massive vector field are not ∂µF

µν = 0, but

∂µFµν +m2Aν = 0 . (1.168)

Applying ∂ν to this relation we conclude that ∂νAν = 0, since ∂ν∂µF

µν is trivially zero. Wecan hence split the latter equation into two

(∂2 +m2)Aµ = 0 , ∂µAµ = 0 . (1.169)

The first equation is a massive Klein-Gordon equation for Aµ with the general solution (1.153)but ωp = (|p|2 + m2)1/2 instead of ωp = |p|. In order to satisfy the second equation we needto impose the condition (1.154), which reduces the number of dofs from four to three. Can wescrap another dof? No we can’t, because the mass term (1.167) breaks gauge invariance andthus we do not have the freedom to use (1.152) and gauge it away. This means that a massivevector field has three physical dofs, one more than a massless one.

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Current Sources

In the discussion of the last two subsections we have always ignored the possible presenceof external sources by simply setting Jµ = 0 by hand. The Lagrangian leading to Maxwell’stheory including a source term JµA

µ has already been given in (1.141) and by applying (1.147),we see that a gauge transformation of such a term introduces an additional piece into theaction:

S → S −∫d4x (∂µf) Jµ = S +

∫d4x f (∂µJ

µ) . (1.170)

As usual we have used integration by parts to arrive at the final answer. This result tells usthat the apparent breaking of gauge invariance by the term JµA

µ can be avoided, if we requirethat

∂µJµ = 0 . (1.171)

In other words the vector field Aµ should couple to a conserved current Jµ. In a fundamentaltheory, the current Jµ should arise from fields, rather than put in by hand as an external source.We conclude the chapter on classical field theory by studying the simplest fundamental theoryof such a kind.

Scalar Electrodynamics And Abelian Higgs Mechanism

The goal of this final subsection is to write down an interesting theory involving a complexscalar ϕ and a vector field Aµ. By interesting we mean that the two fields should interact,which tells us that we want to have a Lagrangian that involves terms with both ϕ and Aµ. Theproblem is now that in general ϕ and Aµ do not transform trivially under a given symmetry.E.g., the complex scalar ϕ transforms as (1.100) under a global U(1), while Aµ obeys the gaugetransformation (1.147). The important point to notice is that both transformations involveonly a single parameter, namely α and f . However, α = const. is space-time independent, whilef = f(x) is a function of the space-time coordinate x. What happens if we generalize (1.100)by allowing the U(1) symmetry to be local rather than global? Meaning that our fields ϕ andϕ∗ should transform as

ϕ→ ϕ′ = eiα(x)ϕ , ϕ∗ → (ϕ′)∗ = e−iα(x)ϕ∗ . (1.172)

Since the scalar potential is polynomial in ϕ∗ϕ it is clearly invariant under these local U(1)transformations. But as already noted after (1.102), for the kinetic term to be invariant, weneed α = const. Explicitly, we have

∂µϕ→ ∂µ(eiα(x)ϕ

)= eiα(x) (∂µ + i∂µα(x))ϕ . (1.173)

Obviously, the term involving ∂µα(x) will cause (∂µϕ∗)(∂µϕ) to transform non-trivially un-

der (1.172). But not all hope is lost since we still have the gauge transformation (1.147) ofthe vector field, which also involves a term with a derivative. So let us see how (∂µ − iAµ)ϕtransforms. One finds

(∂µ − iAµ)ϕ→ eiα [(∂µ + i∂µα)− i(Aµ + ∂µf)]ϕ = eiα [(∂µ − iAµ) + i∂µ(α− f)]ϕ . (1.174)

30

“Eureka!” if we identify α(x) = f(x) then

Dµϕ = (∂µ − iAµ)ϕ , (1.175)

has the correct properties so that

(Dµϕ)∗(Dµϕ)→ (Dµϕ)∗(Dµϕ) , (1.176)

underϕ→ eiαϕ , Aµ → Aµ + ∂µα . (1.177)

In other words the new kinetic term (Dµϕ)∗(Dµϕ) with ∂µ replaced by the covariant derivativeDµ is gauge invariant. Notice that the covariant derivative (1.175) has dimension [Dµ] = 1and involves a “minimal” coupling Aµϕ between the vector field Aµ and the scalar ϕ.

We can now combine our scalar and gauge field Lagrangian to obtain

L = (Dµϕ)∗(Dµϕ)− V (ϕ∗ϕ)− 1

4FµνF

µν , V (ϕ∗ϕ) = V0 +m2ϕ∗ϕ+λ

4(ϕ∗ϕ)2 , (1.178)

which is the Lagrangian of scalar electrodynamics. Notice that in this new theory the role ofAµ is to facilitate the invariance of the scalar field theory under local gauge transformations.In fact, if we would have started from the global U(1) invariant scalar theory with the task offinding its locally U(1) invariant version we would have necessarily been led to introducing agauge field Aµ.

We will now use (1.178) as a toy model to study the spontaneous symmetry breaking of alocal U(1) symmetry. Since the scalar potential is identical to the one in the global case, wecan reuse all our old results. The case m2 ≥ 0 is again not very interesting since the symmetryis unbroken, so let’s focus on the situation with m2 < 0. In the latter case the potential takesthe form of a “Mexican hat” as in Figure 1.4. We parametrize ϕ around the minima in thefollowing way

ϕ =1√2

(v0 + h) eiχ , (1.179)

with v0 given in (1.107). It is straightforward to see that h corresponds to a real massive field,while χ is a Goldstone mode (also a real field). We can now eliminate the Goldstone modefrom ϕ by choosing the specific gauge

ϕ→ ϕ′ = e−iχϕ =1√2

(v0 + h) . (1.180)

Under this transformation, we see from (1.177) that

Aµ → A′µ = Aµ − ∂µχ . (1.181)

Since the Lagrangian (1.178) is gauge invariant, we can as well write it in terms of primedrather than unprimed fields. For the covariant derivative, one finds

(Dµϕ)′ =1√2

(∂µh− i (v0 + h)A′µ

), (1.182)

31

and inserting this expression into (1.178) leads to

L =1

2(∂µh)2 +

1

2A′µA

′µ (v20 + 2v0h+ h2)− V (h)− 1

4F ′µνF

′µν , (1.183)

with

V (h) = V0 +1

4m2v20 −m2h2 +

λ

16

(4v0h

3 + h4). (1.184)

The most striking feature of this result is that the Goldstone mode χ has (magically) disap-peared and we are only left with the fields h and A′µ. So did we loose a dof? No, because thevector field is now no longer massless, but has a mass

mA′ = v0 . (1.185)

It has hence three dofs (as opposed to just two in the massless case), which explains thedisappearance of χ: the Goldstone boson has been “eaten” by the vector field and providesthe additional longitudinal component for the massive A′µ. This feature can also be inferredfrom (1.181). Notice furthermore that the mass of h is given by

mh =√−2m2 , (1.186)

and that (1.183) and (1.184) have a rich structure of interactions: there are couplings of twogauge fields to one and two massive scalars (A′A′h and A′A′h2) as well as cubic and quarticself-couplings of the massive scalar (h3 and h4).

So let’s briefly summarize our main findings: we have learned that a spontaneously brokenlocal (or gauge) symmetry leads to a mass for the associated vector boson and the conversionof the Goldstone boson into the longitudinal component of the gauge field. This is the famousHiggs mechanism that we already alluded to. The massive scalar h is hence called Higgs scalar.The phenomenon described above (but in its generalization to non-abelian gauge groups) isat work in the SM of particle physics. In fact, since the SM symmetry breaking pattern isSU(2)L×U(1)Y → U(1)em, one has three Goldstone bosons that give mass to three electroweakgauge bosons. The masses of these particles are all proportional to the symmetry breakingscale (or vacuum expectation value) v0. A detailed discussion of the breaking of the electroweaksymmetry, while worthwhile, is beyond the scope of this lecture, as it requires the introductionof non-abelian gauge symmetries. You will learn a little bit more in an exercise.

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